Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{k^2 + 8k + 16}{k + 4} \div \dfrac{10k + 40}{k + 2} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{k^2 + 8k + 16}{k + 4} \times \dfrac{k + 2}{10k + 40} $ First factor the quadratic. $z = \dfrac{(k + 4)(k + 4)}{k + 4} \times \dfrac{k + 2}{10k + 40} $ Then factor out any other terms. $z = \dfrac{(k + 4)(k + 4)}{k + 4} \times \dfrac{k + 2}{10(k + 4)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (k + 4)(k + 4) \times (k + 2) } { (k + 4) \times 10(k + 4) } $ $z = \dfrac{ (k + 4)(k + 4)(k + 2)}{ 10(k + 4)(k + 4)} $ Notice that $(k + 4)$ appears twice in both the numerator and denominator so we can cancel them. $z = \dfrac{ \cancel{(k + 4)}(k + 4)(k + 2)}{ 10(k + 4)\cancel{(k + 4)}} $ We are dividing by $k + 4$ , so $k + 4 \neq 0$ Therefore, $k \neq -4$ $z = \dfrac{ \cancel{(k + 4)}\cancel{(k + 4)}(k + 2)}{ 10\cancel{(k + 4)}\cancel{(k + 4)}} $ We are dividing by $k + 4$ , so $k + 4 \neq 0$ Therefore, $k \neq -4$ $z = \dfrac{k + 2}{10} ; \space k \neq -4 $